Question: The lifespans of bears in a particular zoo are normally distributed. The average bear lives $47.6$ years; the standard deviation is $8.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a bear living less than $72.8$ years.
Solution: $47.6$ $39.2$ $56$ $30.8$ $64.4$ $22.4$ $72.8$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $47.6$ years. We know the standard deviation is $8.4$ years, so one standard deviation below the mean is $39.2$ years and one standard deviation above the mean is $56$ years. Two standard deviations below the mean is $30.8$ years and two standard deviations above the mean is $64.4$ years. Three standard deviations below the mean is $22.4$ years and three standard deviations above the mean is $72.8$ years. We are interested in the probability of a bear living less than $72.8$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the bears will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the bears will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $22.4$ years and the other half $({0.15\%})$ will live longer than $72.8$ years. The probability of a particular bear living less than $72.8$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.